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3 years ago

Scarlett needs to order 6,318 binder clips for an upcoming conference. If the binder clips come in

Mathematics

2 answers:

Kryger [21]3 years ago

8 0

Answer: 162

Step-by-step explanation:

Divide 6,318 divided by 39 and you get 162.

Plus why defuk she need that manny binder clip. Does she eat them

Lerok [7]3 years ago

3 0

Answer:

6318 / 39 = 162 boxes.

This question does not really make sense because you ask how many boxes are needed, then say that the clips are coming in 39 boxes.

You might be interested in

Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at

Oksanka [162]

a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively. $\blacksquare$

To learn more on transformation rules, we kindly invite to check this verified question: brainly.com/question/4801277

7 0

3 years ago

Which triangle congruence is this?<br> SSS<br> SAS<br> ASA<br> AAS<br> HL

MrRissso [65]

Answer:

Step-by-step explanation:

The Hypotenuse-Leg Theorem states that two right triangles are congruent if and only if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of the other right triangle.

7 0

3 years ago

The slope of (1, 5), (–3, 0) and (–2, 4), (3, 0)

chubhunter [2.5K]

Answer:

The slope of (1,5) and (-3,0) is 5/4

the slope of (-2,4) and (3,0) is -4/5

Step-by-step explanation:

5 0

4 years ago

Please Help Me!

Rainbow [258]

For the first option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. The range is easily affected by the presence of outliers (data points that are away from other data points). Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Thus, that the range of the two data sets are equal does not mean that the data sets have the same variability. Therefore, the first option is not the correct answer.

For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.

For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.

4 0

3 years ago

Read 2 more answers

Write the expression that represents the phrase: the quotient of 10 and x

Yanka [14]

Answer:

Step-by-step explanation:

$Quotient \ of \ 10 \ and \ x = 10 \div x$

4 0

3 years ago